Question

An object moves along a coordinate line so that its position "s" satisfies the eqn.: s = 4t^2-4t+8. The unit "s" is measured in cm, and "t" is measured in seconds.

1.) Determine velocity of object when t = 1
2.) Determine velocity of object when t = 6
3.) When the velocity is 0, how many seconds have passed?

Answer 1

Is this a calculus class, or a calculus based physics class?

Answer 2

After the questions we have been getting lately, I was beginning to wonder the same thing! (The class is Calculus I)

Answer 3

Calculus and physics are the same thing pretty much. Newton was a physicist who developed calculus to aid in his physics.

Since you are in Calculus, you know that the derivative of position is velocity.

\[s(t)=4t^2-4t+8\]\[s'(t)=v(t)\]Find s'(t)

Answer 4

Velocity as the derivative of position makes sense, think how your change in position over change in time\[\Delta position/\Delta time=distance/time=km/h,\ mi/h, etc.\]

Answer 5

Thank You for your help stamp! I will try to apply your wisdom to the best of my ability.

Answer 6

What did you get for s'(t)=v(t)?

Answer 7

(Still working on it!)

Answer 8

@mylittlepony are you using the power rule? What methods have you learned in lecture for finding a derivative of a function.

Answer 9

Using the power rule, I took the derivative of the position function (as you advised) to yield an equation for velocity.

Position Function: 4t^2-4t+8 = s
Derivative: 8t-4 = s/t

1.) t=1, (s/t)=4

2.) t=6, (s/t)=44

3.) t = 1/2, t/s=0

Thank You so much Stamp!!! (I would have never gotten this question w/o you!)

Answer 10

\[s(t)=4t^2-4t+8\]\[s'(t)=v(t)=8t-4=4(2t-1)\]\[(a).\ Find\ v(1).\]\[v(1)=8(1)-4=4\ cm/s\]\[(b).\ Find\ v(6)\]\[v(6)=8(6)-1=45\ cm/s\]\[(c).\ Find\ t\ when\ v(t)=0\]\[0=8t-4\]\[4(2t-1)=0\]\[(2t-1)=0\]\[v(t)=0\ when\ t=1/2\ s\]

Answer 11

I made an error on (b), it should be\[v(t)=8t-4\]\[v(6)=8(6)-4=44\ cm/s\]